【PAT】A1072 Gas Station【Dijkstra算法】

内容导读

互联网集市收集整理的这篇技术教程文章主要介绍了【PAT】A1072 Gas Station【Dijkstra算法】，小编现在分享给大家，供广大互联网技能从业者学习和参考。文章包含4151字，纯文字阅读大概需要6分钟。

内容图文

A gas station has to be built at such a location that the minimum distance between the station and any of the residential housing is as far away as possible. However it must guarantee that all the houses are in its service range.

Now given the map of the city and several candidate locations for the gas station, you are supposed to give the best recommendation. If there are more than one solution, output the one with the smallest average distance to all the houses. If such a solution is still not unique, output the one with the smallest index number.

Input Specification:

Each input file contains one test case. For each case, the first line contains 4 positive integers: N (≤10^3 ), the total number of houses; M (≤10), the total number of the candidate locations for the gas stations; K (≤10^4?? ), the number of roads connecting the houses and the gas stations; and DS?? , the maximum service range of the gas station. It is hence assumed that all the houses are numbered from 1 to N, and all the candidate locations are numbered from G1 to GM.
Then K lines follow, each describes a road in the format

P1 P2 Dist

where P1 and P2 are the two ends of a road which can be either house numbers or gas station numbers, and Dist is the integer length of the road.

Output Specification:

For each test case, print in the first line the index number of the best location. In the next line, print the minimum and the average distances between the solution and all the houses. The numbers in a line must be separated by a space and be accurate up to 1 decimal place. If the solution does not exist, simply output No Solution.

Sample Input 1:

4 3 11 5
1 2 2
1 4 2
1 G1 4
1 G2 3
2 3 2
2 G2 1
3 4 2
3 G3 2
4 G1 3
G2 G1 1
G3 G2 2

Sample Output 1:

G1
2.0 3.3

Sample Input 2:

2 1 2 10
1 G1 9
2 G1 20

Sample Output 2:

No Solution

题意

给出N个居民点和M个加油站的候选位置以及加油站的服务距离Ds，求离最近居民点最远且平均距离最近的加油站（必须能服务到每一个居民点）。

思路

将题目中1~N映射到0~N-1，G1~GM映射到N~N+M-1，分别以G1~GM为源点做M次Dijkstra求最短路，依次判断并更新求得最终结果。
我使用的是改进版的Dijkstra算法，比《算法笔记》上略快一些。

《算法笔记》对应代码提交结果

改进版Dijkstra算法提交结果

代码

#include <iostream>
#include <vector>
#include <queue>
#include <cstring>
#include <algorithm>
#include <limits.h>
#define MAX_N 1000
#define MAX_K 10000
using namespace std;
struct Edge{
    int to;
    int weight;
    Edge(int to, int weight) : to(to), weight(weight){};
};
struct Node{
    int to;
    int weight;
    Node(int to, int weight) : to(to), weight(weight){};
};
bool operator < (Node a,Node b){
    return a.weight < b.weight;
}
int N, M, K, D;
int getIndex(char s[]){
    int start = 0, index = 0;
    if(s[0] == 'G') start = 1;
    for(int i = start, l = (int)strlen(s); i < l; i++){
        index = index * 10 + (s[i] - '0');
    }
    if(start == 0){
        return index - 1;
    }else{
        return index + N - 1;
    }
}
int bestId = -1;
float bestTotal = 0, bestMin = 0;
void update(int id, int distTo[]){
    float min = -1, total = 0;
    for(int i = 0; i < N; i++){
        if(distTo[i] > D){// 遇到一个距离大于服务半径的居民点，退出
            return;
        }else if(min == -1 || distTo[i] < min){
            min = distTo[i];
        }
        total += distTo[i];
    }
    
    // 根据题意更新答案
    if(bestId == -1 || min > bestMin || (min == bestMin && total < bestTotal)){
        bestTotal = total;
        bestMin = min;
        bestId = id;
    }
}
int main() {
    scanf("%d %d %d %d", &N, &M, &K, &D);
    vector<vector<Edge>> G(N + M);
    char s1[5], s2[5];
    for(int i = 0, weight, u, v; i < K; i++){
        scanf("%s %s %d", s1, s2, &weight);
        u = getIndex(s1);
        v = getIndex(s2);
        G[u].push_back(Edge(v, weight));
        G[v].push_back(Edge(u, weight));
    }
    
    priority_queue<Node> pq;
    bool marked[MAX_N + MAX_K];
    int distTo[MAX_N + MAX_K];
    int l = N + M;
    for(int i = N, u, v; i < l; i++){
        // Dijkstra求最短路
        fill(marked, marked + l, false);
        fill(distTo, distTo + l, INT_MAX);
        pq.push(Node(i, 0));
        distTo[i] = 0;
        while (!pq.empty()) {
            u = pq.top().to;
            pq.pop();
            if(marked[u]) continue;
            for(Edge& e : G[u]){
                v = e.to;
                if(marked[v]) continue;
                if(distTo[u] + e.weight < distTo[v]){
                    distTo[v] = distTo[u] + e.weight;
                    pq.push(Node(v, distTo[v]));
                }
            }
        }
        
        // 更新答案
        update(i - N, distTo);
    }
    
    if(bestId == -1){
        printf("No Solution\n");
    }else{
        printf("G%d\n%.1f %.1f",bestId + 1, bestMin, bestTotal / N);
    }
    return 0;
}

内容总结

以上是互联网集市为您收集整理的【PAT】A1072 Gas Station【Dijkstra算法】全部内容，希望文章能够帮你解决【PAT】A1072 Gas Station【Dijkstra算法】所遇到的程序开发问题。 如果觉得互联网集市技术教程内容还不错，欢迎将互联网集市网站推荐给程序员好友。

内容备注

版权声明：本文内容由互联网用户自发贡献，该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容， 请发送邮件至 gblab@vip.qq.com 举报，一经查实，本站将立刻删除。

内容手机端

---